Property not inherited

A property assumed to pass to subobjects, quotients, or through a chain. It does not.

Would you have fallen for these?

Each is a real option from a previous-year paper. Read it cold and decide before opening it.

• Option B

  f′ must be continuous.

  Why this is wrong, and why it looks right

  For $f(x)=x^{2}\sin(1/x) (x\ne0), f(0)=0: f'(0)=lim_{h\to0}$ h sin(1/h)=0 by the squeeze theorem, but for x≠0, f′(x)=2x sin(1/x)−cos(1/x), whose −cos(1/x) term oscillates without settling as x→0. So f is differentiable everywhere yet f′ is discontinuous at 0.

  From this question

  Let $f : \mathbb{R} \to \mathbb{R}$ be differentiable everywhere. Which of the following are true?

  Part C · free

• Option A

  2

  Why this is wrong, and why it looks right

  Two is the generator count of the infinite cyclic group $\mathbb{Z} ($namely ±1); that count doesn't carry over to the finite quotient $\mathbb{Z}/12\mathbb{Z}$. There, a generator is any k with gcd(k,12)=1, and more than two residues qualify.

  From this question

  The number of generators of the cyclic group $\mathbb{Z}/12\mathbb{Z}$ is

  Part B · free

• Option D

  z·$e^{1/z}$ has a pole at 0.

  Why this is wrong, and why it looks right

  z·$e^{1/z}$ $= z\cdot\sum_{n\ge0}$ $z^{-n}/n$! $= z + 1 + 1/(2z) + 1/(6z^{2}) + \cdots$ still has infinitely many negative powers of z, from the 1/(n!·$z^{n-1})$ terms with n ≥ 2. Here the multiplying factor is z itself, which has a simple zero at 0 rather than being nonvanishing there, so it shifts every exponent in $e^{1/z}$'s Laurent series up by exactly 1 — cancelling only the single n=1 term $(z^{-1}$ becomes the constant 1) while every n≥2 term stays at a negative power. An essential singularity has infinitely many negative-power terms to begin with, and no finite-order zero can cancel all of them, so z·$e^{1/z}$ keeps an essential singularity rather than downgrading to a pole.

  From this question

  Classify the singularity at z = 0 of each function. Which statements are correct?

  Part C · free

• Option D

  $\mathbb{Z}[x]$ is a principal ideal domain.

  Why this is wrong, and why it looks right

  (2,x) cannot be generated by one element: any single generator would have to divide both 2 and x, forcing it to be a unit, yet $(2,x) \ne \mathbb{Z}[x]$. Being a PID does not pass from R to R[x] unless R is a field, and $\mathbb{Z}$ is not one.

  From this question

  Consider the ring $\mathbb{Z}[x]$. Which of the following are true?

  Part C · free

• Option D

  F(x) has at most one real root.

  Why this is wrong, and why it looks right

  False. It reads the root count of f across to F, but a property of a function need not survive antidifferentiation. With $f = x^{3} + x$ and C = −1, SymPy gives F two real roots, $\pm\sqrt{\sqrt{5} - 1}$.

  From this question

  Let f(x) be a cubic polynomial with real coefficients. Suppose that f(x) has exactly one real root and that this root is simple. Which one of the following statements holds for ALL antiderivatives F(x) of f(x)?

  See the full question

  Part B · december 2023 · free

• Option D

  (11/9)X ~ Poisson(11/12)

  Why this is wrong, and why it looks right

  False. Scaling a Poisson variable does not give a Poisson variable: (11/9)X takes values 0, 11/9, 22/9, … which are not non-negative integers, so it cannot have any Poisson distribution. The Poisson family is closed under addition of independent variables, not under multiplication by constants.

  From this question

  Suppose X ~ Poisson(3/4). Then which of the following statements is true?

  See the full question

  Part B · december 2023 · free

• Option D

  at the centre of the rectangle

  Why this is wrong, and why it looks right

  A centred peak is a feature of specific initial data, e.g. the first Dirichlet eigenfunction $\sin(\pi{}x/L)$, not a property every solution must have. The maximum principle only guarantees the max sits somewhere on the parabolic boundary, not at any particular point of it.

  From this question

  Let u solve the heat equation u_t = u_xx on 0 < x < L, 0 < t ≤ T, with continuous data. By the maximum principle, the maximum of u over the closed rectangle is attained

  Part B · free

• Option D

  X is connected

  Why this is wrong, and why it looks right

  For any $1/n_{0}$, an interval around it short enough to miss its neighbours makes {$1/n_{0}$} itself clopen in X, so X splits into two nonempty relatively open pieces — X is totally disconnected. Sitting inside the connected line $\mathbb{R}$ does not force a subset to be connected.

  From this question

  Let X = {$1/n : n \in \mathbb{N}$} ∪ {0} with the usual metric from $\mathbb{R}$. Which of the following are true?

  Part C · free

• Option B

  It may have eigenvalues

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option C

  The analogous Fredholm equation (fixed integration limits) is also uniquely solvable for every $\lambda$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option B

  If |G| = 2, then Aut(G × G) is abelian.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

• Option C

  If G is the group of complex numbers under addition, then Aut(G) is abelian.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

• Option C

  If $G_{1}$ is normal in $G_{2}$ and $G_{2}$ is normal in G, then $G_{1}$ is normal in G.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

• Option A

  $\mathbb{R}[x]/(f)$ is a field.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option A

  S is a principal ideal domain

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

• Option B

  $S/(y^{2} + x^{2})$ is a unique factorization domain

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

• Option C

  Every rearrangement of $\sum{}a_{n}$ converges to the same sum

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option B

  Every quotient of a Hausdorff space is Hausdorff.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option A

  If f is continuous on the set S, then f is continuous on the set $\mathbb{R}$ \ S

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2024

• Option B

  If g is continuous, then f is continuous on the set S

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2024

• Option A

  If f(a) is a unit in S for every non-zero element a ∈ R, then S is a field

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

• Option B

  If f(a) is a unit in S for every non-zero element a ∈ R, then f(R) is a field

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

• Option D

  19

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option D

  (T − 2I)(T − 3I) is a nilpotent operator.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

• Option A

  If X and Y are compact, Y is Hausdorff and f is onto, then X is also Hausdorff.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option B

  If X is an infinite compact set and f is a homeomorphism from X to f(X) (where f(X) is given the subspace topology), then Y is compact.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option C

  If X is Hausdorff and f is onto, then Y is Hausdorff.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option C

  connected whenever it has more than one point

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option C

  uniformly to a continuous function

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option B

  closed but not open

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option D

  neither open nor closed

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option C

  f must have a zero.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option A

  Every quotient ring of R is a principal ideal domain

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

• Option B

  There exists a quotient ring S of R and an ideal I ⊆ S which is not principal

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

• Option D

  simple

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option A

  $\mathbb{R}$ ∖ $\mathbb{Q}$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option D

  If each $f_{n}$ is differentiable and $(f_{n})_{n}\ge_{1}$ converges to f uniformly, then f is differentiable.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

The counterexamples that settle these

One example is enough to kill a false statement. These are the standard ones for the claims above.

• Closed and bounded ⇒ compact
• Bounded ⇒ totally bounded
• Continuous on a bounded interval ⇒ bounded
• Uniformly continuous ⇒ Lipschitz
• Differentiable ⇒ continuously differentiable
• Pointwise limit of continuous functions is continuous

Other mistakes that cost marks